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The problem provides the function \( g(u, \lambda) \) defined as:

\[
g(u, \lambda) = (u - 1)(u - 2) + \lambda \sin(u)
\]

We are asked to analyze the solutions for \( g(u, \lambda) = 0 \) when \( \lambda = 0 \), and determine which solutions persist when \( \lambda \) becomes nonzero.

### Step 1: Analyze the case when \( \lambda = 0 \)
When \( \lambda = 0 \), the function simplifies to:

\[
g(u, 0) = (u - 1)(u - 2)
\]

The solutions to \( g(u, 0) = 0 \) are:

\[
(u - 1)(u - 2) = 0
\]

Thus, \( u = 1 \) and \( u = 2 \) are the two solutions when \( \lambda = 0 \).

### Step 2: Analyze the persistence of solutions when \( \lambda \neq 0 \)
Now, let's consider \( \lambda \neq 0 \), i.e., the function becomes:

\[
g(u, \lambda) = (u - 1)(u - 2) + \lambda \sin(u)
\]

We want to determine whether the solutions \( u = 1 \) and \( u = 2 \) persist when \( \lambda \) is small but nonzero.

- **At \( u = 1 \):**

Substituting \( u = 1 \) into the equation:

\[
g(1, \lambda) = (1 - 1)(1 - 2) + \lambda \sin(1) = 0 + \lambda \sin(1)
\]

Since \( \sin(1) \neq 0 \), we have:

\[
g(1, \lambda) = \lambda \sin(1)
\]

For small nonzero \( \lambda \), this expression is nonzero, which means that \( u = 1 \) is no longer a solution when \( \lambda \neq 0 \).

- **At \( u = 2 \):**

Similarly, substituting \( u = 2 \) into the equation:

\[
g(2, \lambda) = (2 - 1)(2 - 2) + \lambda \sin(2) = 0 + \lambda \sin(2)
\]

Since \( \sin(2) \neq 0 \), we have:

\[
g(2, \lambda) = \lambda \sin(2)
\]

Again, for small nonzero \( \lambda \), this expression is nonzero, implying that \( u = 2 \) is no longer a solution when \( \lambda \neq 0 \).

### Conclusion:
Neither of the solutions \( u = 1 \) and \( u = 2 \) persists when \( \lambda \) becomes nonzero, as both solutions lead to nonzero values in the equation for small nonzero \( \lambda \).

Would you like a deeper explanation or more examples?

Here are five related questions for further exploration:
1. How can we find approximate solutions for small nonzero \( \lambda \)?
2. What happens to the behavior of the function as \( \lambda \) increases?
3. How can we analyze the stability of the solutions as \( \lambda \) changes?
4. Can we use perturbation methods to find corrections to the solutions for small \( \lambda \)?
5. What are the geometric interpretations of the function \( g(u, \lambda) \) for different values of \( \lambda \)?

**Tip:** Small parameter expansions (like perturbation theory) are a powerful tool to find approximate solutions when parameters are close to zero.

Let g : R × R → R be given by g(u, λ) = (u - 1)(u - 2) + λ sin(u). When λ = 0, the equation has two solutions, namely, u = 1 and u = 2. Which solutions persist as λ becomes nonzero?

This problem explores the function g(u, λ) = (u - 1)(u - 2) + λ sin(u) and investigates which solutions persist as λ moves from zero to nonzero values. Key concepts include perturbation theory and trigonometric functions, ideal for undergraduate-level studies.

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